# Thread: Prove there is a number a

1. ## Prove there is a number a

Suppose f: D --> R is a function whose domain contains [a,b] for some a,b $\displaystyle \in$ R amd f is integrable on [a,b]. suppose a < c < b.

1. Suppose A is a set of real number that is bounded above. Prove that for any $\displaystyle \epsilon$ > 0 there is a number a $\displaystyle \in$A Such that supA-a < $\displaystyle \epsilon$.
(Hint: Prove by contradiction. If there is an $\displaystyle \epsilon$ >0 such that supA-a >= $\displaystyle \epsilon$ for all a $\displaystyle \in$ A then what does this say about the number supA - $\displaystyle \epsilon$ ?)

2. Suppose there exists $\displaystyle \epsilon>0$ such as $\displaystyle \sup A-a\geq \epsilon,\forall a\in A\Rightarrow \sup A-\epsilon \geq a,\forall a\in A\Rightarrow \exists b<\sup A:b\geq a,\forall a\in A$. Hence, $\displaystyle \sup A$ isn't the lower upper bound.

3. Originally Posted by Abu-Khalil
Suppose there exists $\displaystyle \epsilon>0$ such as $\displaystyle \sup A-a\geq \epsilon,\forall a\in A\Rightarrow \sup A-\epsilon \geq a,\forall a\in A\Rightarrow \exists b<\sup A:b\geq a,\forall a\in A$. Hence, $\displaystyle \sup A$ isn't the lower upper bound.
Can you explain what we are proving here, supA - a < $\displaystyle \epsilo$ implies it is a least upper bound or not a least upper bound?

4. Originally Posted by 450081592
Suppose f: D --> R is a function whose domain contains [a,b] for some a,b $\displaystyle \in$ R amd f is integrable on [a,b]. suppose a < c < b.

1. Suppose A is a set of real number that is bounded above. Prove that for any $\displaystyle \epsilon$ > 0 there is a number a $\displaystyle \in$A Such that supA-a < $\displaystyle \epsilon$.
(Hint: Prove by contradiction. If there is an $\displaystyle \epsilon$ >0 such that supA-a >= $\displaystyle \epsilon$ for all a $\displaystyle \in$ A then what does this say about the number supA - $\displaystyle \epsilon$ ?)
This is really quite simple my good man. Suppose for a second that there existed some $\displaystyle \varepsilon>0$ such that no element of $\displaystyle A$ was greater than $\displaystyle \sup\text{ }A-\varepsilon$. Then, it must be an upper bound of $\displaystyle A$ but since $\displaystyle \text{sup }A-\varepsilon<\text{sup }A$ this contradicts $\displaystyle \sup\text{ }A$ being the least upper bound.

5. Originally Posted by Drexel28
This is really quite simple my good man. Suppose for a second that there existed some $\displaystyle \varepsilon>0$ such that no element of $\displaystyle A$ was greater than $\displaystyle \sup\text{ }A-\varepsilon$. Then, it must be an upper bound of $\displaystyle A$ but since $\displaystyle \text{sup }A-\varepsilon<\text{sup }A$ this contradicts $\displaystyle \sup\text{ }A$ being the least upper bound.
Ya I am clear now, thanks a lot